We derive the gradient of \eqref{eq:roll-tpr-at-fpr-gaussian} with respect to
$f_\theta(\mathbf{x}_i)$, instantiating the general structure~\eqref{eq:roll-gradient}
for the Gaussian case. Throughout, $\tau = \hat{F}_0^{-1}(1-\alpha) =
\mu_0 + \sigma_0\sqrt{2}\,\operatorname{ierf}(1-2\alpha)$.

\paragraph{Case $y_i = 1$.}

The score $f_\theta(\mathbf{x}_i)$ enters through $\mu_1$ and $\sigma_1$.
Differentiating $\hat{F}_1(\tau) = \frac{1}{2}\bigl[1+\operatorname{erf}\!\bigl(\frac{\tau-\mu_1}{\sigma_1\sqrt{2}}\bigr)\bigr]$:

\[
  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(1-\alpha))}{\partial \mu_1} =
  -\frac{1}{\sigma_1\sqrt{2\pi}}\exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
\]

\[
  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(1-\alpha))}{\partial \sigma_1} =
  \frac{\mu_1 - \tau}{\sigma_1^2\sqrt{2\pi}}\exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
\]

Combined with $\frac{\partial \mu_1}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{|\mathcal{B}_1|}$
and $\frac{\partial \sigma_1}{\partial f_\theta(\mathbf{x}_i)} = \frac{f_\theta(\mathbf{x}_i)-\mu_1}{|\mathcal{B}_1|\sigma_1}$,
the chain rule gives:

\begin{equation}
  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(1-\alpha))}{\partial f_{\theta}(\mathbf{x}_i)}
  = -\frac{1}{|\mathcal{B}_1|\,\sigma_1\sqrt{2\pi}}
    \exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
    \left(1 + \frac{(\tau - \mu_1)(f_\theta(\mathbf{x}_i) - \mu_1)}{\sigma_1^2}\right)
  \label{eq:gauss-grad-y1}
\end{equation}

\paragraph{Case $y_i = 0$.}

The score $f_\theta(\mathbf{x}_i)$ affects the loss only through $\tau$. By the chain rule:

\[
  \frac{\partial \hat{F}_1(\tau)}{\partial f_\theta(\mathbf{x}_i)}
  = \underbrace{\frac{\partial \hat{F}_1(\tau)}{\partial \tau}}_{\text{PDF of class 1 at }\tau}
    \cdot \frac{\partial \tau}{\partial f_\theta(\mathbf{x}_i)}
\]

The first factor is the Gaussian PDF at $\tau$:
\[
  \frac{\partial \hat{F}_1(\tau)}{\partial \tau}
  = \frac{1}{\sigma_1\sqrt{2\pi}}\exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
\]

For the second factor, since $\tau = \mu_0 + \sigma_0\sqrt{2}\,\operatorname{ierf}(1-2\alpha)$:
$\frac{\partial\tau}{\partial\mu_0}=1$ and
$\frac{\partial\tau}{\partial\sigma_0}=\frac{\tau-\mu_0}{\sigma_0}$.
Combined with $\frac{\partial\mu_0}{\partial f_\theta(\mathbf{x}_i)}=\frac{1}{|\mathcal{B}_0|}$
and $\frac{\partial\sigma_0}{\partial f_\theta(\mathbf{x}_i)}=\frac{f_\theta(\mathbf{x}_i)-\mu_0}{|\mathcal{B}_0|\sigma_0}$:
\[
  \frac{\partial\tau}{\partial f_\theta(\mathbf{x}_i)}
  = \frac{1}{|\mathcal{B}_0|}\!\left(1+\frac{(\tau-\mu_0)(f_\theta(\mathbf{x}_i)-\mu_0)}{\sigma_0^2}\right)
\]

Therefore:
\begin{equation}
  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(1-\alpha))}{\partial f_\theta(\mathbf{x}_i)}
  = \frac{1}{|\mathcal{B}_0|\,\sigma_1\sqrt{2\pi}}
    \exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
    \left(1 + \frac{(\tau - \mu_0)(f_\theta(\mathbf{x}_i) - \mu_0)}{\sigma_0^2}\right)
  \label{eq:gauss-grad-y0}
\end{equation}

Substituting \eqref{eq:gauss-grad-y1} and \eqref{eq:gauss-grad-y0} into \eqref{eq:roll-gradient} yields \eqref{eq:gauss-grad-combined}.